Introduction to energy problem - did I do this right?

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SUMMARY

The discussion centers on a physics problem involving a bullet penetrating a block of wood, where the bullet's mass is 25g, its initial velocity is 350 m/s, and the opposing force is 50,000N. The user attempted to apply the work-energy principle, concluding that the bullet penetrates 0.03m. However, a participant named Chet pointed out that while the user's methodology is correct, the problem itself is flawed, as average force cannot directly determine displacement without knowing the time of impact.

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Homework Statement


The bullet strikes a block of wood which exerts, on average, a force of 50,000N opposing the motion of the bullet. How far does the bullet penetrate?

Mass of bullet = 25g
Initial velocity of bullet = 350 m/s
Final velocity of bullet = 0 m/s

Homework Equations


Ei+work=Ef
K = ½mv^2
Energy = Force * Displacement

The Attempt at a Solution


Please excuse my attempt of it's egregiously wrong, I was just introduced to energy yesterday; this is one of the first problems I'm doing.

Anyways, because the final velocity of the bullet is zero, the bullet must have lost all of its energy, making Ef=0 J. The bullet initially had kinetic energy, making Ei=K. Thus, my new equation is
K+work = 0 J
½mv^2+work=0 J

I also know that the problem gave me how many Newtons oppose the object, but I need to find out how much energy it used. Since Energy = Force*Displacement, I can rewrite the equation as:
½mv^2+(F*Δx) = 0 J
So I solved it below...
½(.025kg)(350 m/s)^2+(-50000N*Δx) = 0 J
Δx=.03m

Did I do this problem right? If not, what did I do wrong?
 
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I didn't check your arithmetic, but you methodology is correct.

Chet
 
It's the right answer according to what the questioner seems to expect you to do, but the question is actually wrong.
Knowing the average force you cannot deduce the distance, only the time. Your method effectively assumes a constant force. Average force is defined as ##\Delta##momentum/##\Delta##time. In general, this gives a different number from ##\Delta##KE/displacement.
 

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